Find constants A and B such that \(\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\)

for all x such that\( x\neq -1 and x\neq 2. \)Give your answer as the ordered pair (A,B).

unknowledgeable Jul 30, 2020

#1**0 **

If you let x go infinity, then you get 0 = A + B.

If you let x = 0, then you get -7/2 = A/(-2) + B.

The solution to this system is then A = 7/3, B = -7/3, so (A,B) = (7/3,-7/3).

Guest Jul 30, 2020

#2**+1 **

Hmm, not quite! if you let x go to infinity then you get 0 = 0.

Let x go to 0 is fine. Probably better to let x go to 1 rather than infinity for another choice though.

Alan Jul 30, 2020

#3**+2 **

wasting lots of sticky notes i finally got it

\(\frac{x+7}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}\) now mulitply (x-2)(x+1) on both sides

\(x+7=Ax+A+Bx-2B\) trying an example of x, i use 0 which gives

\(7=A-2B\) and do another that is 4

\(11=5A+2B\) the I add both equations together

\(18=6A\) which leaves us with A=3 plugging this into the equation

\(7=3-2B\) and so B equals -2 leaving us with

the ordered pair for (A,B) as \(3,-2\)

unknowledgeable Jul 30, 2020